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Vera_Pavlovna [14]
4 years ago
13

If you want to change the pitch of a sound, you have to change the ___

Chemistry
2 answers:
aleksandrvk [35]4 years ago
5 0

Answer: If you want to change the pitch of a sound, you have to change the <em>frequency.</em>

Explanation:

Pitch is the quality of the sound to judge it has higher or lower.The frequency of the sound wave is perceived as pitch.Pitch and frequency of the sound are related to each other

  • If want to increase the pitch we have to increase the frequency of the sound.
  • If want to decrease the pitch we have to decrease the frequency of the sound.

If you want to change the pitch of a sound, you have to change the <em>frequency.</em>

Firdavs [7]4 years ago
4 0

if you are wanting to change the pitch of a certain sound you ahve and will need to change the frequency of that sound.

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Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2
quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done

p_{ext} = external pressure

V_1 = initial volume of gas

V_2 = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

w=-nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 25^oC=273+25=298K

V_1 = initial volume of gas

V_2 = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(2.5atm)\times (10-1)L

w=-22.5L.atm=-22.5\times 101.3J=-2279.25J

(2) A free isothermal expansion from 1 L to 100 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})

w=-11409.6J

(3) A reversible isothermal expansion from 0.5 L to 4 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})

w=-5151.97J

(4) A reversible isothermal expansion from 0.5 L to 40 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})

w=-10856.8J

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(0.5atm)\times (100-1)L

w=-49.5L.atm=-49.5\times 101.3J=-5014.35J

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0
3 years ago
Copper metal reacts with nitric acid according the equation below 3 Cu (s) +8 HNO 3(aq) -...m+3 Cu^ 2+ (aq)+2NO (g) +6 NO 3 (aq)
Maksim231197 [3]

Answer:

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3 years ago
Explain in an essay how the food chain works? ​
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Explanation:

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A sample of gas contains 0.1100 mol of N2(g) and 0.3300 mol of H2(g) and occupies a volume of 20.5 L. The following reaction tak
worty [1.4K]

Answer:

20.5 L

Explanation:

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N_2_{(g)}+3H_2_{(g)}->2NH_3_{(G)}

-We apply the combined gas law to determine the resultant volume of the ammonia gas formed:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\\\#But\\\\\frac{P_1}{T_1}=\frac{P_2}{T_2}\\\\\therefore V_1=V_2

Since, the temperature and pressure remain constant, the volume of the ammonia gas formed is equal to the volume of the reactants.

Hence, the volume after the reaction is 20.5 L

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4 years ago
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